/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 5 Iced tea On Tuesday, the bottles... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Iced tea On Tuesday, the bottles of Arizona Iced Tea filled in a plant were supposed to contain an average of 20ounces of iced tea. Quality control inspectors selected 50bottles at random from the day’s production. These bottles contained an average of 19.6 ounces of iced tea. Identify the population, the parameter, the sample, and the statistic.

Short Answer

Expert verified

The sample is of 50bottles and population minimum temperature becomes associated parameter. The statistics is 19.6 ounces and parameter is population mean weight of iced tea.

Step by step solution

01

Given information

We need to identify the population, the parameter, the sample, and the statistic.

02

Simplify

Individuals that seek to collect information are present in the population.
A sample is a portion of the population from whom data was gathered.
The sample is seen to be the 50 bottles chosen at random from the day's production, whereas the population must be all bottles from the day's production.
A statistic is a descriptive measure for a sample, whereas a parameter is a descriptive measure for a population.
The average of 19.6ounce is based on 50bottles in the sample, and so 19.6ounces indicate the sample mean weight of the iced tea, implying that 19.6 is a statistic.
The population minimum temperature becomes the associated parameter.
Statistic :

x¯=sample mean weight of the iced tea =19.6 ounces
Parameter : μ= population mean weight of the iced tea

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that you have torn a tendon and are facing surgery to repair it. The orthopedic surgeon explains the risks to you. Infection occurs in 3%of such operations, the repair fails in 14%, and both infection and failure occur together 1%of the time. What is the probability that the operation is successful for someone who has an operation that is free from infection?

a. 0.8342

b. 0.8400

c. 0.8600

d. 0.8660

e. 0.9900

The math department at a small school has 5teachers. The ages of these teachers are 23,34,37,42,58. Suppose you select a random sample of 4teachers and calculate the sample minimum age. Which of the following shows the sampling distribution of the sample minimum age?

a.

b.

c.

d.

e. None of these

Dem bones (2.2) Osteoporosis is a condition in which the bones become brittle due to the loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in a standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD score that is 2.5standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and gender roughly follow a Normal distribution.

a. What percent of healthy young adults have osteoporosis by the WHO criterion?

b. Women aged 70to 79are, of course, not young adults. The mean BMD in this age group is about-2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis?

Social scientists are interested in the association between high school graduation rate (HSGR, measured as a percent) and the percent of U.S. families living in poverty (POV). Data were collected from all 50 states and the District of Columbia, and a regression analysis was conducted.

The resulting least-squares regression line is given by POV∧=59.2-0.620(HSGR) POV^=59.2-0.620(HSGR) with r2=0.802r2=0.802. Based on the information, which of the following is the best interpretation for the slope of the least-squares regression line?

a. For each 1% increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.896 .

b. For each 1 % increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.802.

c. For each 1 % increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.620.

d. For each 1 % increase in the percent of families living in poverty, the graduation rate is predicted to decrease by approximately 0.802.

e. For each 1 % increase in the percent of families living in poverty, the graduation rate is predicted to decrease by approximately 0.620.

IQ tests The Wechsler Adult Intelligence Scale (WAIS) is a common IQ test for adults. The distribution of WAIS scores for persons over 16 years of age is approximately Normal with mean 100 and standard deviation $15 .

a. What is the probability that a randomly chosen individual has a WAIS score of 105 or greater?

b. Find the mean and standard deviation of the sampling distribution of the average WAIS score x-x¯for an SRS of 60 people. Interpret the standard deviation.

c. What is the probability that the average WAIS score of an SRS of 60 people is 105 or greater?

d. Would your answers to any of parts (a), (b), or (c) be affected if the distribution of WAIS scores in the adult population was distinctly non-Normal? Explain your reasoning.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.